Revision Questions (1 Viewer)

[shafqat]

Dynamics

1. a) a = -16pi^2x
b) v^2 = 16pi^2(400 – x^2)
c) 6sqrt11

2.a) 2sqrt6 km/s
b) 28000/3sqrt6 sec

3. c) 1/2sqrt kg . log3 secs

 

[shafqat]

Binomial

1.a) 112
b) -65

2. x = 0, (5 + sqrt10) / 2, (5 - sqrt10) / 2

3.a) 25 choose 10. 7^15. (5x) ^10
b) 25 choose 15 . 7^10. 10^15

4. 1/(n+1)

5. 0, n is odd
(-1)^n/2. n choose n/2

 

[shafqat]

Integration

1. 4/15 (sqrt2 + 1)

2. 1/3

3. (6 + pi – 3sqrt3)/6

4. 2/sqrt3 tan^-1 (2tan @/2 + 1/ sqrt3) + c

5. 16/3 log |x-5/x-3| + c

6. -2log|x-1| + 3log(x^2 + x + 2) + c

7. 1/9 x^3 (3lnx – 1) + c

8. 1/6 (2pi – 3sqrt3)

9. sqrt( x^2 + 2x + 5) + log( x + 1 + x^2 + 2x + 5)

10. sinx – 1/3(sinx)^3

11. 1/5 log6

12. xtan^-1x – ½log(1+ x^2)

13. 2/sqrt7 [ tan^-1 sqrt7/5)]

14. -1/2 (e^pi + 1)

15. b) pi^2/4

16. 1/sqrt2 + ½. log (sqrt2 + 1)

 

[niall045]

thankyou, answers are fine by me.

 

[LaCe]

With De Moivres, could I get some answers for question 3 d) and e), just want to check.

3.c) since w^k is a solution of z^9-1=0
(z-1)(w^8+w^7+...+w+1)=0
since z cannot equal 1
w^8+....+w=-1

EDIT: wait, i have done this question before, it takes a while, does anyone have a quick method for this or do u just have to grind it out with trig?

 

[..:''ooo]

ssafquat thanx for the files, do u have any more?
could u scan all the other review sheets (if u can be bothered)
thanks

 

[KFunk]

De Moivre, volumes, recursion

6. a) 10! (pi^10 / 10! – pi^8/8! + pi^6/6! – pi^4/4! + pi^2/2! – 2)

Do you have the second page of le volumes/recursion sheet? I'm curious to see the question that goes with that answer.

 

[shafqat]

Do you have the second page of le volumes/recursion sheet? I'm curious to see the question that goes with that answer.

Sure thing.

 

[KFunk]

Very prompt. Cheers mate.

 

[Vampire]

hmmmm i'm having the same problems as LaCe....

 

[FinalFantasy]

With De Moivres, could I get some answers for question 3 d) and e), just want to check.

3.c) since w^k is a solution of z^9-1=0
(z-1)(w^8+w^7+...+w+1)=0
since z cannot equal 1
w^8+....+w=-1

EDIT: wait, i have done this question before, it takes a while, does anyone have a quick method for this or do u just have to grind it out with trig?

3c)to prove: w+w²+w³+...+w^8=-1
first prove 1+w+w²+w³+....+w^8=0
LHS=1(1-w^9)\(1-w)
but w^9=1
.: LHS=1(1-1)\(1-w)=0
.: 1+w+w²+w³+....+w^8=0
hence w+w²+w³+...+w^8=-1

dats my way, dunno if it's a "quick method" or not
plz make comments if it's not rite or not good

 

[FinalFantasy]

d)w+w²+w^4=cos(2pi\9)+cos (4pi\9)+cos(8pi\9)+i(sin(2pi\9)+sin(4pi\9)+sin(8pi\9))
let @=(2pi\9)

1+w+w²+w³+...+w^8=0
equate real and imaginary parts
1+cos@+cos2@+cos3@+...+cos8@=0
cos@+cos2@+cos4@=-1-cos8@-cos7@-cos6@-cos5@-cos3@
cos@+cos2@+cos4@=-1-(cos@)-(cos2@)+(1\2)-(cos4@)+(1\2)
cos@+cos2@+cos4@=-cos@-cos2@-cos4@
2(cos@+cos2@+cos4@)=0
.: cos@+cos2@+cos4@=0
back to:
w+w²+w^4=cos(2pi\9)+cos (4pi\9)+cos(8pi\9)+i(sin(2pi\9)+sin(4pi\9)+sin(8pi\9))
w+w²+w^4=0+i(....)
.: it's pure imaginary

i think dis covers part of E) as well..
i know it's not an elegant solution but o well..

 

[LaCe]

I was talking about question 3e)

Thanks mate for d), its not all that long, however my way for e) takes a while with adding cos values repeatedly till u get a product